parsing output

[prad87]

i need to parse the following output and get all the lines representing a particular row

the output is :

<resultSet>
<row>
<column xsi:type="Key">
<id>145</id>
<policy className="Meter" id="144"/>
<name>key1</name>
<description>key</description>
<type>SESSION</type>
</column>
</row>
<row>
<column xsi:type="Key">
<id>150</id>
<policy className="Meter" id="149"/>
<name>key2</name>
<description>first key</description>
<type>SESSION</type>
</column>
</row>
</resultSet>

Now I need to get the row based on the <id>
So I know the id say 145, now I need to print out all the elements in that row.
Please help me out with this.I need to do this in shell script


Thanks in advance

[gerard4143]

Sounds like a homework problem - "I need to do this in shell script"

[prad87]

do u have an idea on how to do it?

[scathefire]

what language you gonna write it in?

[nmset]

After looking at your previous posts, you don't seem to be a guy doing homeworks.

The following script is not a general purpose xml bash parser, it just does what you want. C++/C, Java... have got full XML processors and parsers for real flexibility and productivity.

The idea is to write all elements of a row on one line, then select the required line by the id tag, and finally present the data as in the original file.

#!/bin/bash
IN="/var/tmp/xmlforbash.xml"
OUT="/var/tmp/toRows.xml"
rm -f "$OUT"
row=""
while read line
do
  [ "$line" != "<resultSet>" ] && row="$row$line"
  if [ "$line" = "</row>" ]; then
    echo "$row" >> "$OUT"
    row=""
  fi
done < "$IN"
grep "<id>145</id>" "$OUT" | sed 's/></>\n</g'
#rm -f "$OUT"